Problem: Drazil likes heap very much. So he created a problem with heap:

There is a max heap with a height h implemented on the array. The details of this heap are the following:

This heap contains exactly 2^h - 1 distinct positive non-zero integers. All integers are distinct. These numbers are stored in the array a indexed from 1 to 2^h-1. For any 1 < i < 2^h, a[i] < a[\left ⌊{i/2}\right ⌋].

Now we want to reduce the height of this heap such that the height becomes g with exactly 2^g-1 numbers in heap. To reduce the height, we should perform the following action 2^h-2^g times:

Choose an index i, which contains an element and call the following function f in index i:

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Note that we suppose that if a[i]=0, then index i don't contain an element.

After all operations, the remaining 2^g-1 element must be located in indices from 1 to 2^g-1. Now Drazil wonders what's the minimum possible sum of the remaining 2^g-1 elements. Please find this sum and find a sequence of the function calls to achieve this value.

Input

The first line of the input contains an integer t (1 ≤ t ≤ 70 000): the number of test cases.

Each test case contain two lines. The first line contains two integers h and g (1 ≤ g < h ≤ 20). The second line contains n = 2^h-1 distinct positive integers a[1], a[2], …, a[n] (1 ≤ a[i] < 2^{20}). For all i from 2 to 2^h - 1, a[i] < a[\left ⌊{i/2}\right ⌋].

The total sum of n is less than 2^{20}.

Output

For each test case, print two lines.

The first line should contain one integer denoting the minimum sum after reducing the height of heap to g. The second line should contain 2^h - 2^g integers v_1, v_2, …, v_{2^h-2^g}. In i-th operation f(v_i) should be called.

Example

Input


2
3 2
7 6 3 5 4 2 1
3 2
7 6 5 4 3 2 1


Output


10
3 2 3 1
8
2 1 3 1

Solution (in JAVA):

Solution:
//package com.company;


import java.io.*;
import java.math.*;
import java.util.*;

public class Main {
    public static class Task {
        public class ValueWithIndex {
            int val, index;
            public ValueWithIndex(int v, int i) {
                this.val = v;
                this.index = i;
            }
        }
        int[] values = new int[1 << 20];
        boolean[] remain = new boolean[1 << 20];
        int[] sVal = new int[1 << 20];
        List<ValueWithIndex> mergeWithMin(List<ValueWithIndex> l, List<ValueWithIndex> r, int v) {
            int lIdx = 0, rIdx = 0;
            while (lIdx < l.size() && l.get(lIdx).val <= v) lIdx++;
            while (rIdx < r.size() && r.get(rIdx).val <= v) rIdx++;
            List<ValueWithIndex> s = new ArrayList<>();
            while (lIdx < l.size()) {
                while (rIdx < r.size() && r.get(rIdx).val <= l.get(lIdx).val) {
                    s.add(r.get(rIdx));
                    rIdx++;
                }
                s.add(l.get(lIdx));
                lIdx++;
            }
            while (rIdx < r.size()) {
                s.add(r.get(rIdx));
                rIdx++;
            }
            return s;
        }
        long sum = 0;
        public List<ValueWithIndex> search(int index, int h, int g) {
            if (index * 2 + 1 < (1 << h)) {
                List<ValueWithIndex> left = search(index * 2, h, g);
                List<ValueWithIndex> right = search(index * 2 + 1, h, g);
                int L = sVal[index * 2];
                int R = sVal[index * 2 + 1];
                List<ValueWithIndex> m = mergeWithMin(left, right, Math.max(L, R));
                m.add(new ValueWithIndex(values[index], index));
                if (index < (1 << g)) {
                    sVal[index] = m.get(0).val;
                    remain[m.get(0).index] = true;
                    sum += sVal[index];
                }
                return m;
            }
            List<ValueWithIndex> me = new ArrayList<>();
            me.add(new ValueWithIndex(values[index], index));
            return me;
        }
        public void solve(Scanner sc, PrintWriter pw) throws IOException {
            int T = sc.nextInt();
            while (T-- > 0) {
                int h = sc.nextInt();
                int g = sc.nextInt();
                for (int i = 0; i < 1 << h; i++) {
                    values[i] = sVal[i] = 0;
                    remain[i] = false;
                }
                for (int i = 0; i < (1 << h) - 1; i++) {
                    values[i + 1] = sc.nextInt();
                }
                sum = 0;
                search(1, h, g);
                pw.println(sum);
                for (int i = (1 << h) - 1; i >= 1; i--) {
                    if (!remain[i]) pw.print(i + " ");
                }
                pw.println();
            }
        }
    }

    static long TIME_START, TIME_END;

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
//        Scanner sc = new Scanner(new FileInputStream("input"));
        PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));
//        PrintWriter pw = new PrintWriter(new FileOutputStream("output"));
//        int t = (1 << 20) / 3;
//        pw.println(t);
//        for (int i = 0; i < t; i++) {
//            pw.println("2 1");
//            pw.println("1 2 3");
//        }
        Runtime runtime = Runtime.getRuntime();
        long usedMemoryBefore = runtime.totalMemory() - runtime.freeMemory();
        TIME_START = System.currentTimeMillis();
        Task t = new Task();
        t.solve(sc, pw);
        TIME_END = System.currentTimeMillis();
        long usedMemoryAfter = runtime.totalMemory() - runtime.freeMemory();
        pw.close();
        System.err.println("Memory increased: " + (usedMemoryAfter - usedMemoryBefore) / 1000000);
        System.err.println("Time used: " + (TIME_END - TIME_START) + ".");
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(FileReader s) throws FileNotFoundException {
            br = new BufferedReader(s);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}